The relation between torque and angular acceleration is

Explanation

(i) Consider a rigid body rotating about a fixed axis. A point mass m in the body will execute a circular motion about a fixed axis as shown in figure.

(ii) Let `vec(F)` be a tangential force acting on the point mass produces the torque for rotation and `vec(r)` be the position vector of the point mass.
(iii) The torque produced by the force on the point mass m about the axis of rotation is written as,
`{:(tau = r F Sin 90 = r F" "[because sin 90^(@) =1]),(tau = r ma" "[because (F = ma)]),(tau = r m ralpha = mr^(2) alpha" "[because (a = r alpha)]):}`
`tau = (mr^(2))alpha" "...(1)`
(iv) For all particles of the body, `mr^(2) = sum m_(i) r_(i)^(2)`
(v) Therefore, `tau = (sum m_(i)r_(i)^(2))a`
`tau = I alpha`
(vi) Where, `I = sum m_(i)r_(i)^(2)` moment of inertia of the rigid body.
(vii) In vector form, `vec(tau) = I vec(alpha)`